Problem: $\lim_{x\to\infty}\dfrac{(\ln(x))^3}{\ln(4x)}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{3}{4}$ (Choice C) C $3$ (Choice D) D $\infty$
Solution: $\lim_{x\to\infty} (\ln(x))^3=\infty$ and $\lim_{x\to\infty} \ln(4x)=\infty$, so $\lim_{x\to\infty}\dfrac{(\ln(x))^3}{\ln(4x)}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{(\ln(x))^3}{\ln(4x)} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[(\ln(x))^3\right]}{\dfrac{d}{dx}[\ln(4x)]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{3(\ln(x))^2\cdot\dfrac{1}{x}}{\dfrac{4}{4x}} \\\\ &=\lim_{x\to\infty}3(\ln(x))^2 \\\\ &=\infty \end{aligned}$ Note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[(\ln(x))^3\right]}{\dfrac{d}{dx}[\ln(4x)]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{(\ln(x))^3}{\ln(4x)}=\infty$.